a) \(x^2+2x+1-y^2=\left(x+1\right)^2-y^2=\left(x+1-y\right)\left(x+1+y\right)\)

b) \(x^2+4x+3=x^2+4x+4-1=\left(x+2\right)^2-1=\left(x+1\right)\left(x+3\right)\)

c) \(4x^2-9y^2=\left(4x-9y\right)\left(4x+9y\right)\)

d) \(x^3-27y^3=\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)

a)\(x^2+2x+1-y^2=\left(x+1\right)^2-y^2=\left(x+y-1\right)\left(x-y+1\right)\)

b)\(x^2+4x+3=\left(x+1\right)\left(x+3\right)\)

c)\(4x^2-9y^2=\left(2x-3y\right)\left(2x+3y\right)\)

d)\(x^3-27y^3=\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)

Mk ko ghi laj đề nha

\(=\left(17x^4:4x^2\right)-\left(5x^3:4x^2\right)+\left(2x^2:4x^2\right)\)

\(=\frac{17}{4}x^2-\frac{5}{4}x+\frac{2}{4}\)

\(=\frac{17}{4}x^2-\frac{5}{4}x+\frac{1}{2}\)

MK KO GHI LAJ ĐỀ NHA

\(=\left(17x^4:4x^2\right)-\left(5x^3:4x^2\right)+\left(2x^2:4x^2\right)\)

\(=\frac{17}{4}x^2-\frac{5}{4}x+\frac{1}{2}\)

2Mg + O2 => 2MgO

n(MgO) = 8/40 = 0,2 (mol)

=> n(Mg) = 0,2 => m(Mg) = 0,2.24 = 4,8g

n(O2) = 0,2/2 = 0,1 => V(O2) = 2,24 l

( trên kia phải là điều kiện tiêu chuẩn chứ nhỉ )

Ta có P=a^2+4b^2=(a^2-4ab+4b^2) +4ab

                             =(a-2b)^2+4ab=5^2+4*4=41

Vậy P=41 tại a-2b=5, ab=4

Ta có: (a-2b)^2 +4ab = 5^2 + 4.4

           a^2 -4ab + 4b^2 +4ab = 25+16

           a^2 + 4b^2 = 41

Vậy P =41

Đặt \(P=y^2+3xy-12x^2\)

\(4P=\left[\left(2y\right)^2+2.2.3xy+\left(3x\right)^2\right]-57x^2\)

\(4P=\left(2y+3x\right)^2-\left(\sqrt{57}x\right)^2\)

\(4P=\left(2y+3x-\sqrt{57}x\right).\left(2y+3x+\sqrt{57}x\right)\)

\(P=\frac{1}{4}.\left(2y+3x-\sqrt{57}x\right).\left(2y+3x+\sqrt{57}x\right)\)

Tham khảo nhé~

Bài 2:

\(\left(5x+1\right)^2-\left(2xy-3\right)^2\)

\(=25x^2+10x+1-\left(2xy-3\right)^2\)

\(=25x^2+10x+1\left(4x^2y^2-12xy+9\right)\)

\(=25x^2+10x+1-4x^2y^2+12xy-9\)

\(=25x^2-4x^2y^2+10x+12xy-8\)

Bài 2: 

\(\left(x-1\right)\left(x^2+x+1\right)=x^2\left(x-9\right)+2x+6\)

\(=x^3-1=x^3-9x^2+2x+6\)

\(=x^3-9x^2+2x+6=x^3-1\)

\(=x^3-9x^2+2x+6+1=x^3-1+1\)

\(=x^3-9x^2+2x+7=x^3\)

\(=x^3-9x^2+2x+7-x^3=x^3-x^3\)

\(=-9x^2+2x+7=0\)

\(\Rightarrow x=-\frac{7}{9};x=1\)

Ta có: \(a^3+b^3+c^3=3abc\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(2a^2+2b^2+2c^2-2ab-2bc-2ca\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left[\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\right]=0\)

\(\Leftrightarrow\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\end{cases}}\)

*) Xét a + b + c = 0 => \(\hept{\begin{cases}-a=b+c\\-b=c+a\\-c=a+b\end{cases}}\)

\(\Rightarrow A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}\cdot\frac{b+c}{c}\cdot\frac{c+a}{a}=\frac{\left(-c\right).\left(-a\right).\left(-b\right)}{b.c.a}=-1\)

*) Xét \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Rightarrow a=b=c}\)

\(\Rightarrow A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

Vậy A = -1 hoặc A = 8